2.1.4d-e - Acid-base titrations

2.1.4d-e - Acid-base titrations

Acid-base titration is a practical method for finding an unknown concentration by reacting a measured volume with a solution whose concentration is known accurately. The chemistry is simple neutralisation, but the result is only trustworthy if the volumes, standard solution and mole-ratio calculation are handled carefully. This lesson links the practical technique to the calculation route, so the numbers mean something rather than becoming a set of disconnected steps.

Purpose and key terms

A titration compares two reacting solutions. One solution has a known concentration and the other is being found. A measured fixed volume of one solution is placed in a conical flask, then the other solution is added from a burette until the reaction is just complete.

Standard solution

A standard solution is a solution with an accurately known concentration.

Titre

A titre is the volume delivered from the burette in a titration. It is calculated from final burette reading - initial burette reading.

The endpoint is the point where the indicator shows a colour change. The equivalence point is the chemical point where the reacting amounts are exactly in the mole ratio in the balanced equation. A good titration method chooses and uses an indicator so that the observed endpoint is close enough to the equivalence point for an accurate concentration calculation.

The fixed volume placed in the conical flask is often called an aliquot. In school and college titrations this is commonly 25.00 cm^3 measured using a volumetric pipette. Repeated accurate titres should be close together; these are called concordant titres. In many practical methods, titres within 0.10 cm^3 are treated as concordant, but always follow the tolerance stated in the method or question.

Purpose and key terms Continued

The most important idea is that each apparatus choice protects one part of the calculation. The volumetric flask fixes the final volume of the standard solution. The pipette fixes the aliquot volume. The burette measures the variable volume needed to reach the endpoint. The balanced equation gives the mole ratio.

Preparing a standard solution

To prepare a standard solution from a solid, the aim is to put a known amount of substance into an accurately known final volume.

A careful sequence is:

  1. Calculate the mass of solid needed.
  2. Weigh the solid accurately, often by difference using a weighing bottle or boat.
  3. Dissolve the solid in a beaker using a small volume of deionised water.
  4. Transfer the solution to a volumetric flask using a funnel.
  5. Rinse the beaker, stirring rod and funnel into the volumetric flask so no dissolved solid is left behind.
  6. Add deionised water until the bottom of the meniscus is on the calibration mark at eye level.
  7. Stopper the flask and invert it several times to make the solution uniform.

The final volume must be the volume in the volumetric flask, not the volume of water first used in the beaker. The beaker stage is for dissolving and quantitative transfer; the volumetric flask stage is what gives the accurate final concentration.

Amount, concentration and volume

n=cVn = cV

where n is amount in mol, c is concentration in mol dm^{-3}, and V is volume in dm^3.

Worked example: prepare 250.0 cm^3 of 0.100 mol dm^-3 sodium carbonate solution, Na2CO3.

First convert the volume:

250.0 cm3=0.2500 dm3250.0\ \text{cm}^3 = 0.2500\ \text{dm}^3

Find the amount needed:

n=cV=0.100×0.2500=0.02500 moln = cV = 0.100 \times 0.2500 = 0.02500\ \text{mol}

For Na2CO3, Mr = 106.0.

m=nMr=0.02500×106.0=2.650 gm = nM_r = 0.02500 \times 106.0 = 2.650\ \text{g}

So the method should use about 2.650 g of Na2CO3 and make the dissolved solid up to exactly 250.0 cm^3 in a volumetric flask. A balance reading to 2 or 3 decimal places may be expected depending on the apparatus used; keep the final calculated concentration to an appropriate number of significant figures.

Preparing a standard solution Continued

Two errors are especially damaging. If some solid is lost during transfer, the actual concentration is lower than intended. If the flask is filled above the mark, the concentration is also lower because the same amount is spread through too large a volume.

Carrying out the titration

The burette, pipette and conical flask each have a different job.

The burette delivers a measured variable volume. Rinse it with the solution it will contain before filling it; otherwise water left inside dilutes the solution. Remove the funnel after filling, because drops falling from the funnel during the titration would change the volume without being included in the initial reading.

The volumetric pipette transfers a fixed aliquot, usually 25.00 cm^3. Rinse it with the solution it will transfer. Use a pipette filler, never mouth pipetting. Let the pipette drain naturally; do not blow out the last drop unless the pipette is designed for that.

The conical flask holds the aliquot and indicator. It does not need to be dry, and it may be rinsed with deionised water during the titration because extra water changes concentration but not the amount of substance already in the flask. Add only a few drops of indicator, then swirl the flask as the burette solution is added.

A practical titration usually begins with a rough titre. After that, repeat accurate titres, adding the titrant quickly at first and dropwise near the endpoint. A sharp endpoint is a colour change that remains after swirling. Read the burette at eye level and record readings to a sensible precision, usually two decimal places for a burette scale read to the nearest 0.05 cm^3.

Worked example: choosing a mean titre.

TitrationInitial / cm^3Final / cm^3Titre / cm^3
Rough0.0025.4025.40
10.1024.8024.70
20.0024.6024.60
30.2024.8024.60

The rough titre is not used in the mean. Titres 2 and 3 are concordant because they agree exactly. Titre 1 is close, but if a question clearly has two best concordant titres, use those. The mean selected titre is:

24.60+24.602=24.60 cm3\frac{24.60 + 24.60}{2} = 24.60\ \text{cm}^3

Carrying out the titration Continued

Risk management is part of the method. Acids and alkalis can be irritant or corrosive depending on concentration, so wear eye protection, clean spills promptly, and keep the burette below eye level when filling. Phenolphthalein and methyl orange are common indicators, but this lesson only needs the practical idea that the indicator should give a clear endpoint.

Structured titration calculations

Most titration calculations use the same core route:

  1. Convert the titre from cm^3 to dm^3.
  2. Use n = cV to find moles of the solution with known concentration.
  3. Use the balanced equation to convert moles of one reactant into moles of the other.
  4. Use the pipette volume or total solution volume to find the required concentration.

Worked example: 25.00 cm^3 of hydrochloric acid is pipetted into a conical flask. It requires 23.60 cm^3 of 0.100 mol dm^-3 sodium hydroxide for neutralisation.

Equation:

HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

Find moles of NaOH:

V=23.60 cm3=0.02360 dm3V = 23.60\ \text{cm}^3 = 0.02360\ \text{dm}^3 n(NaOH)=cV=0.100×0.02360=0.002360 moln(\text{NaOH}) = cV = 0.100 \times 0.02360 = 0.002360\ \text{mol}

The mole ratio HCl : NaOH is 1 : 1, so:

n(HCl)=0.002360 moln(\text{HCl}) = 0.002360\ \text{mol}

This amount of HCl was in 25.00 cm^3:

25.00 cm3=0.02500 dm325.00\ \text{cm}^3 = 0.02500\ \text{dm}^3 c(HCl)=nV=0.0023600.02500=0.0944 mol dm3c(\text{HCl}) = \frac{n}{V} = \frac{0.002360}{0.02500} = 0.0944\ \text{mol dm}^{-3}

Structured titration calculations Continued

The balanced equation is not decoration. It is the step that changes the measured moles of one solution into the reacting moles of the other. A common error is to assume every acid-base titration has a 1 : 1 ratio; this only works when the equation says so.

Less-structured titration problems

In less-structured questions, the calculation route is the same, but you must decide which volume or scaling step matters. Label each amount as you go. This prevents the most common mistake: calculating the amount in the 25.00 cm^3 aliquot, then forgetting to scale up to the full volumetric flask.

A useful route is:

  1. Identify the known concentration and the titre.
  2. Calculate moles of the known solution used in the titre.
  3. Use the equation to find moles of the unknown in the aliquot.
  4. If the aliquot came from a larger solution, multiply by the dilution factor.
  5. Use the question's target: concentration, mass, percentage purity or molar mass.

If the target is percentage purity, first use the titration to find the mass of the pure acid or base present. Then compare that with the mass of the original sample:

percentage purity=mass of pure substancemass of sample×100\text{percentage purity} = \frac{\text{mass of pure substance}}{\text{mass of sample}} \times 100

For example, if titration data show that a 1.50 g sample contains 1.20 g of pure acid, the percentage purity is (1.20 / 1.50) x 100 = 80.0%.

Worked example: a 1.260 g sample of a solid diprotic acid, H2A, is dissolved and made up to 250.0 cm^3 in a volumetric flask. 25.00 cm^3 portions of this solution require a mean titre of 21.00 cm^3 of 0.150 mol dm^-3 NaOH(aq). Calculate the molar mass of H2A.

Equation:

H2A(aq)+2NaOH(aq)Na2A(aq)+2H2O(l)\text{H}_2\text{A(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{A(aq)} + 2\text{H}_2\text{O(l)}

Moles of NaOH in the titre:

V=21.00 cm3=0.02100 dm3V = 21.00\ \text{cm}^3 = 0.02100\ \text{dm}^3 n(NaOH)=0.150×0.02100=0.003150 moln(\text{NaOH}) = 0.150 \times 0.02100 = 0.003150\ \text{mol}

The equation shows 1 mol H2A reacts with 2 mol NaOH:

n(H2A in 25.00 cm3)=0.0031502=0.001575 moln(\text{H}_2\text{A in 25.00 cm}^3) = \frac{0.003150}{2} = 0.001575\ \text{mol}

The full solution is 250.0 cm^3, which is 10 times the aliquot:

n(H2A in 250.0 cm3)=0.001575×10=0.01575 moln(\text{H}_2\text{A in 250.0 cm}^3) = 0.001575 \times 10 = 0.01575\ \text{mol}

Now use mass = amount x molar mass, rearranged:

Mr=mn=1.2600.01575=80.0 g mol1M_r = \frac{m}{n} = \frac{1.260}{0.01575} = 80.0\ \text{g mol}^{-1}

Less-structured titration problems Continued

For unfamiliar acids and bases, do not guess the formula from the name unless the question gives enough information. Use the balanced equation, formula or stated basicity/acidity provided in the question to get the mole ratio.

Quality and common errors

Good titration answers show both chemical method and mathematical control. Use cm^3 for recorded volumes, but convert to dm^3 before using concentration in mol dm^-3. Avoid unsupported units such as ml or l in formal working; use cm^3 and dm^3 consistently.

Record burette readings and titres with an appropriate number of decimal places. A titre written as 24.6 cm^3 may lose precision compared with 24.60 cm^3 if the apparatus supports two decimal places. In a calculation, give the final answer to a sensible number of significant figures, normally guided by the data in the question.

When choosing a mean titre, do not average all results automatically. Exclude the rough titre and any non-concordant result. If several accurate titres are concordant, average the selected concordant set and state what you used.

Common practical and calculation errors include:

  • using the trial titre in the mean
  • forgetting cm^3 to dm^3, so the answer is 1000 times too large or too small
  • using the wrong mole ratio from the balanced equation
  • treating the aliquot amount as the amount in the whole volumetric flask
  • forgetting a mass conversion such as g to mg, if the question asks for it
  • over-shooting the endpoint and accepting a titre that is too high
  • failing to rinse the burette or pipette with the solution they will contain

A reliable titration answer is a chain: accurate solution preparation, careful endpoint technique, selected concordant mean, n = cV, balanced-equation ratio, then any required scaling or conversion.